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I was reading an electronics book and came upon this circuit.

enter link description here

I am having a difficult time understanding why the voltage across D1 is 0.6V. I understand that that the reason is because the input signal's pp voltage is lower than 0.6v C and R1 make a differentiator then D2 rectifies that but none of that would work if the signal is less than 0.6v. I just don't understand why D1 is at 0.6V and how that makes D2 be higher than 0.6V. I was going to post a diagram but I don't have enough reputation.

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  • \$\begingroup\$ I am not sure that I understand exactly where you are having difficulty, so I would like to ask some questions if I may... If the circuit were reduced to just R3 and D1 from 5V supply to ground, do you understand why it is okay to assume that D1 is on the drop across it is 0.6V? (It sounds like you get that part.) Next, add in R1, D2 and R2 (but leave out the capacitor). Is it clear to you why it still must be assumed that (a) diode D1 is on (with 0.6V across it), (b) that diode D2 is off and that (c) current is only flowing through R3 and D1? I'm running out of chars so I'll stop here for now \$\endgroup\$ – Adam P Apr 2 '11 at 0:28
  • \$\begingroup\$ I understand until that point but I don't see why that would make the voltage at the node connecting C, D1, R1 higher than 0.6V so that D2 is on. Assuming that we have a square wave signal with a peak voltage of 0.4V, shouldn't that node be at 0.4V? \$\endgroup\$ – ergodicsum Apr 2 '11 at 6:19
  • \$\begingroup\$ I am going to need some extra space to write so I will post a new answer to continue this discussion. It's tough without drawing and redrawing and annotating the circuit but I'll try my best to explain. \$\endgroup\$ – Adam P Apr 2 '11 at 17:00
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The short answer is that this is the "magic" of capacitive coupling. The signal voltage is superimposed on top of the node "bias" voltage (which is 0.6V). Therefore the voltage at C,D2,R1 is 0.6V + Vin, not just Vin. And therefore the diode D2 must turn on, have a 0.6V drop across it (assuming D2 and D1 identical), and the output voltage must then be basically equal to Vin. It is important to note that just the AC part of Vin gets through the cap. The DC part is blocked, if there is a DC part.

The longer answer goes like this:

In your response to my comment, you wrote "C, D1, R1" but I think you meant "C, D2, R1". So I will continue with that assumption.

First, note the node D1, R1, R3 will basically be clamped to 0.6V at all times. This will not change no matter what the input signal is.

Second, remember that the I-V relationship for a capacitor is

$$i_c = C\frac {dV} {dt}$$

What does this mean? It means that if your input voltage source changes, then the voltage across the cap changes. Therefore you have a nonzero $\frac {dV} {dt}$. Therefore you will have a nonzero current. (It's easy to get confused about signs at this point... bear with me) This is the differentiator action that you mentioned in your question.

So let's assume an increase in input voltage which would cause a current to flow "through" the cap into the circuit (to the right). Where does the current go? It can either go through R1 or D2 or both.

We started with the assumption that D2 is off, so let's first assume that the current flows only through the resistor R1. That causes a voltage to be developed across R1. So the voltage at C,D2,R1 becomes 0.6V + I*R1. Here is kind of the key part for you to see, I think... see that the voltage at that node must have an 0.6V offset? That is because the change in input voltage causes a current to flow through the capacitor, not a voltage. So the current causes an additional voltage to be developed across the resistor R1, which adds to the voltage at that node due to the diode D1.

Okay, so continuing on, this means that the voltage across D2 would be greater than 0.6V, since the cathode ("line side") of the diode is presently at 0V. If the diode forward voltage is 0.6V, then we have an extra voltage (equal to I*R1) that would cause the diode D2 to turn on--which would mean that the diode would present a very low resistance compared to R1, which means current would flow through the diode D2, also.

So, our initial assumption that the current flows only downward through R1 must be incorrect--instead, the current must also flow through D2 and then through R2. Note that the current that flows through D2 is then limited by the 10K resistance R2, so it will indeed be small compared to the current that flows through R1 (by a factor of about 1/10, in fact). But regardless of the amount of current that flows, the output voltage still effectively "follows" the voltage at C,D2,R1, just shifted down by one diode drop (0.6V). EDIT: except of course for negative input swings, in which case D2 blocks. But I think that part is already understood.

That came out pretty lengthy. I hope it was clear. It's tough to talk about circuits without a chalkboard or something to draw on...

EDIT: I reread your comment, and you mention a square wave input. In the case of the square wave, the voltage at C,D2,R1 is not going to look like a square wave. It is going to look like a series of positive and negative spikes. This is due to the differentiating action of the capacitor. The square wave signal is "constant in the short term". It just switches polarity periodically. It is during those switching times that a signal will appear at the C,D2,R1 node, because only during switching is $\frac{dV}{dt}$ nonzero. This is all assuming several "idealities" which I will not list out at this time, in order to keep it simple. It's easier to think about if the input signal were a sine wave (easier for me, anyway).

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  • \$\begingroup\$ Thanks for your answer this really helped understand things. \$\endgroup\$ – ergodicsum Apr 2 '11 at 20:44
  • \$\begingroup\$ @ergodicsum You're welcome, glad I could help. \$\endgroup\$ – Adam P Apr 4 '11 at 2:47
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I'd say your confusion is justified. When the input signal is constant or varying near ground or above, D1 is forward biased, so it's at 0.6 V. However, if the input signal spikes massively downward, such that the voltage between C and R1 is below -5 V, D1 will be reverse biased, so it will act like an open circuit.

For large negative spikes, R1 and R3 act like a voltage divider. With a large enough spike, the voltage across D1 could be as large a negative voltage as you like (until R1 and/or R3 burn out or C explodes).

Of course, I'd assume that the point of the circuit is to handle small positive signals, so in normal operation, you'd just assume the diode was forward biased and thus at 0.6 V.

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The voltage across D1 is 0.6V because of the diode forward voltage drop ($V_{d}$). The forward voltage drop of a diode is almost constant whatever the current passing through the diode so they have a very steep characteristic:

enter image description here

The Volt across D2 is about 0.6V or 0.7V too.

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  • \$\begingroup\$ The problem is not why the voltage drop is 0.6V, it's why the input signal doesn't drop 0.6V across D2. The interesting thing here is the coupling, not $V_d$. \$\endgroup\$ – Kevin Vermeer Apr 2 '11 at 20:07
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This kind of circuit is used in amplitude detectors which (assuming appropriate diodes) can work with signals from kilohertz up to several gigahertz. They give output voltage that is proportional to the amplitude of the input.

If you wire this circuit without $R_1$, $D_1$ and $R_3$ you will get a detector with the nice property that the output current is proportional to the square of the input voltage. This means that if you know the load resistance (which is most likely $50\,\Omega$) you can make a direct power measurement with just a linear scale ($P = {V^2 \over R}). Unfortunately this only holds if the input is below 0.6 V (or 0.3 V for Schottky diodes which are frequently used in such circuits) and the output current is very low (see the diode I-V characteristics below and the pictures in this document).

Diode I-V characteristics(picture by Glen Leinweber)

$C$ is very small (100pF) so you can treat is as a open circuit when considering DC voltages and as a short at high frequencies (it works like a $100\,\Omega$ resistor at 15MHz). This means that adding $R_1$, $D_1$ and $R_3$ will add some polarizing voltage (and current) to the $D_2$ diode. $R_3$ and $D_1$ form a so called shunt voltage regulator. Because of the highly nonlinear characteristics of the diode it's output voltage is largely independent of both the supply voltage and the current through $R_3$ (within some limits).

The output from this regulator is fed through $R_1$ (a small inductor would be even better here) to get some current flowing through $D_2$. This (thanks to $C$) shifts the average input voltage and that results in a change in the operating conditions of $D_2$. At higher voltages a diode permits more current to flow. Also the rate of change of the current with applied voltage is higher (recall the characteristics). This improves the sensitivity and makes it easy to measure thanks to the higher overall output current.

You could do this with a normal resistive divider but this would make the circuit sensitive to temperature and diode parameter variation. OTOH making $D_1$ and $D_2$ the same (in fact two diodes made from one silicon die are the best) will make the output from the shunt regulator follow any changes in the characteristics of $D_2$ (because of temperature, humidity or other stuff) protecting you from over- or under-polarizing.

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    \$\begingroup\$ so .. R1, D1 and R3 are basically just a biasing network for D2, so that the input signals don't have to overcome the usual forward diode drop in D2 on their own? \$\endgroup\$ – JustJeff Apr 2 '11 at 1:11
  • \$\begingroup\$ I understand the voltage divider part and I see why it is 0.6V but I don't understand what you mean by adding polarizing voltage. I get that the voltage at the node joining C R_1 D_2 should be higher than 0.6V but I don't see why \$\endgroup\$ – ergodicsum Apr 2 '11 at 6:14
  • \$\begingroup\$ @user3688 I tried to expand my answer. Let me know what if something remains unclear. \$\endgroup\$ – jpc Apr 2 '11 at 10:54
  • \$\begingroup\$ @JustJeff Yup. They also have some effect on input impedance (lowering it from $10$ to $1\,\mathrm{k\Omega}$). \$\endgroup\$ – jpc Apr 2 '11 at 10:56
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R3 and D1 form a voltage divider with 0.6V in the midpoint. The 0.6V is given by forward voltage of the diode and R3 is there to limit the current through the diode. This voltage is then applied through resistor R1 in between C and D2 as a biasing voltage to make sure that D2 will stay always open, given that input voltage is smaller that the bias.

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